Question

At a certain clinic, 75% of the patients with hamstring injuries who are treated with e-stim (electrical muscle stimulation) return to play their sport within two weeks. Three randomly selected hamstring patients who were treated at this clinic are interviewed.

Part A) What is the probability that all 3 returned to play within two weeks?

Part B) What is the probability that none of the 3 returned to play within two weeks?

Part C) What is the probability that at least one of the 3 returned to play within two weeks?

Answer 1

Solution:
P(patients with hamstring injuries who are treated with the e-stim return to play their sport within two weeks) = 0.75
No. of selected hamstring patients = 3
Here we will use the binomial distribution to calculate probabilities because all three randomly selected hamstring patients who were treated at this clinic interviewed are independent of each other.
P(X = n |N,p) = NCn*(p^n)*((1-p)^(N-n))
Solution(a)
we need to calculate the probability that all 3 returned to play within two weeks
P(X=3 |3,0.75) = 3C3*(0.75^3)*(0.25)^0 = 0.4219
So there is 42.19% probability that all 3 returned to play within two weeks.
Solution(b)
We need to calculate the probability that none of the 3 returned to play within two weeks
P(X=0 |3,0.75) = 3C0*(0.75^0)*(0.25)^3 = 0.0156
So there is 1.56% probability that none of the 3 returned to play within two weeks.
Solution(c)
We need to calculate the probability that at least one of the 3 returned to play within two weeks
P(X>=1 |3,0.75) = 1 - P(X=0) = 1 - 0.0156 = 0.9844
So there is 98.44% probability that at least one of the 3 returned to play within two weeks.